Two parallel plate capacitors with fixed plates are connected to two batteries, as shown in the figure. The plate separation is identical for both capacitors. The plates are rectangular, having a width b and lengths l₁ and l₂ . The left half of the dielectric slab has a dielectric constant K₁ while the right half has K₂ . Neglecting all friction, determine the ratio of the emf of the left battery to that of the right

Step-by-step solution

Let the separation between the plates of both capacitors be d and the width of the plates be b . The capacitance of a parallel plate capacitor with a dielectric slab partially inserted up to a length x is given by: C = ₀ b (l - x) d + K ₀ b x d When the capacitor is connected to a battery of emf V , the force exerted on the dielectric slab is: F = 1 2 V^2 dC dx Differentiating C with respect to x , we get: dC dx = ₀ b (K - 1) d Thus, the force pulling the dielectric into the capacitor is: F = ₀ b V^2 (K - 1) 2d For